A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio $1: 1 : 8: 27: 27$. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to :

We have 1 big cube that gets melted and reformed into 5 smaller cubes.

The 5 cubes have volume ratio $1:1:8:27:27$

We need to find how much MORE surface area the 5 cubes have compared to the original.

---

Let the original cube have volume $= V$

Total ratio parts $= 1 + 1 + 8 + 27 + 27 = 64$

The 5 smaller cubes have volumes:

• Cube 1: $\frac{V}{64}$
• Cube 2: $\frac{V}{64}$
• Cube 3: $\frac{8V}{64} = \frac{V}{8}$
• Cube 4: $\frac{27V}{64}$
• Cube 5: $\frac{27V}{64}$

---

For any cube with volume $v$, the side length $= \sqrt[3]{v}$

Original cube side $= \sqrt[3]{V}$

The 5 smaller cubes have sides:

• Side₁ $= \sqrt[3]{\frac{V}{64}} = \frac{\sqrt[3]{V}}{4}$
• Side₂ $= \frac{\sqrt[3]{V}}{4}$
• Side₃ $= \sqrt[3]{\frac{V}{8}} = \frac{\sqrt[3]{V}}{2}$
• Side₄ $= \sqrt[3]{\frac{27V}{64}} = \frac{3\sqrt[3]{V}}{4}$
• Side₅ $= \frac{3\sqrt[3]{V}}{4}$

---

Surface area of any cube $= 6 \times (\text{side})^2$

Original cube surface area $= 6(\sqrt[3]{V})^2$

Sum of 5 cubes' surface areas:

$= 6\left[\left(\frac{\sqrt[3]{V}}{4}\right)^2 + \left(\frac{\sqrt[3]{V}}{4}\right)^2 + \left(\frac{\sqrt[3]{V}}{2}\right)^2 + \left(\frac{3\sqrt[3]{V}}{4}\right)^2 + \left(\frac{3\sqrt[3]{V}}{4}\right)^2\right]$

$= 6(\sqrt[3]{V})^2\left[\frac{1}{16} + \frac{1}{16} + \frac{1}{4} + \frac{9}{16} + \frac{9}{16}\right]$

$= 6(\sqrt[3]{V})^2\left[\frac{2 + 4 + 18}{16}\right]$

$= 6(\sqrt[3]{V})^2 \times \frac{24}{16} = 6(\sqrt[3]{V})^2 \times 1.5$

---

Percentage increase $= \frac{\text{New} - \text{Original}}{\text{Original}} \times 100\%$

$= \frac{1.5 \times 6(\sqrt[3]{V})^2 - 6(\sqrt[3]{V})^2}{6(\sqrt[3]{V})^2} \times 100\%$

$= \frac{0.5 \times 6(\sqrt[3]{V})^2}{6(\sqrt[3]{V})^2} \times 100\% = 0.5 \times 100\% = 50\%$